Question

8%20%5Csum_%7Bk%20%3D%201%7D%5E%7Bn%7D%20%5Cfrac%7B%20%7Bk%7D%5E%7B2%7D%20%7D%7B%20%7B2k%7D%5E%7B2%7D%20-%202nk%20%2B%20%7Bn%7D%5E%7B2%7D%20%7D%20%5Cright%29%20%5Cleft%28%20%5Csum_%7Bk%20%3D%201%7D%5E%7Bn%7D%20%5Cfrac%7B%20%7Bk%7D%5E3%7D%7B%20%7B3k%7D%5E%7B2%7D%20-%203nk%20%2B%20%7Bn%7D%5E%7B2%7D%20%7D%20%5Cright%29%7D%20" id="TexFormula1" title=" \displaystyle \rm\lim_{n \to \infty } \sqrt[n]{ \left( \sum_{k = 1}^{n} \frac{ {k}^{2} }{ {2k}^{2} - 2nk + {n}^{2} } \right) \left( \sum_{k = 1}^{n} \frac{ {k}^3}{ {3k}^{2} - 3nk + {n}^{2} } \right)} " alt=" \displaystyle \rm\lim_{n \to \infty } \sqrt[n]{ \left( \sum_{k = 1}^{n} \frac{ {k}^{2} }{ {2k}^{2} - 2nk + {n}^{2} } \right) \left( \sum_{k = 1}^{n} \frac{ {k}^3}{ {3k}^{2} - 3nk + {n}^{2} } \right)} " align="absmiddle" class="latex-formula">​ ​

Answer 1

Rewrite the sums as

$\displaystyle S_2 = \sum_{k=1}^n \frac{k^2}{2k^2 - 2nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{2k^2}{n^2} - \frac{2k}n + 1}$

and

$\displaystyle S_3 = \sum_{k=1}^n \frac{k^2}{3k^2 - 3nk + n^2} = \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{3k^2}{n^2} - \frac{3k}n + 1}$

Now notice that

$\displaystyle \lim_{n\to\infty} \frac{S_2}n = \int_0^1 \frac{x^2}{2x^2 - 2x + 1} = \frac12$

and

$\displaystyle \lim_{n\to\infty} \frac{S_3}n = \int_0^1 \frac{x^2}{3x^2 - 3x + 1} = \frac{9 + 2\pi\sqrt3}{27}$

and the important point here is that $\frac{S_2}n$ and $\frac{S_3}n$ converge to constants. For any real constant a, we have

$\displaystyle \lim_{n\to\infty} \frac{\ln(an)}n = 0$

Rewrite the limit as

$\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = \lim_{n\to\infty} \exp\left(\ln\left(\sqrt[n]{S_2 \times S_3}\right)\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln(S_2) + \ln(S_3)}n\right) \\\\ = \exp\left(\lim_{n\to\infty} \frac{\ln\left(n \times \frac{S_2}n\right) + \ln\left(n \times \frac{S_3}n\right)}n\right)$

Then

$\displaystyle \lim_{n\to\infty} \sqrt[n]{S_2 \times S_3} = e^0 = \boxed{1}$

A plot of the limand for n = first 1000 positive integers suggests the limit is correct, but convergence is slow.